Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 100010;
int tot, u[N], v[2 * N], pre[2 * N];
void add_b(int a, int b) {
  v[tot] = b;
  pre[tot] = u[a];
  u[a] = tot++;
}
int dp[N][20][11], de[N], fa[N][20];
void add_p(int a, int b) {
  if (dp[a][1][0] < 10) {
    dp[a][1][0]++;
    dp[a][1][dp[a][1][0]] = b;
  }
}
void merge_dp(int a, int b, int c) {
  int l = 1, r = 1, k = 1;
  while (l <= dp[a][c][0] && r <= dp[b][c][0] && k <= 10) {
    if (dp[a][c][l] <= dp[b][c][r]) {
      dp[a][c + 1][k++] = dp[a][c][l];
      l++;
    } else {
      dp[a][c + 1][k++] = dp[b][c][r];
      r++;
    }
  }
  while (l <= dp[a][c][0] && k <= 10) {
    dp[a][c + 1][k++] = dp[a][c][l];
    l++;
  }
  while (r <= dp[b][c][0] && k <= 10) {
    dp[a][c + 1][k++] = dp[b][c][r];
    r++;
  }
  dp[a][c + 1][0] = k - 1;
}
void dfs(int x, int p) {
  de[x] = de[p] + 1;
  fa[x][0] = x;
  fa[x][1] = p;
  for (int i = 2; i < 20; i++)
    if (1 << (i - 1) >= de[x])
      break;
    else {
      merge_dp(x, fa[x][i - 1], i - 1);
      fa[x][i] = fa[fa[x][i - 1]][i - 1];
    }
  int w = u[x];
  while (w != -1) {
    if (v[w] == p) {
      w = pre[w];
      continue;
    }
    dfs(v[w], x);
    w = pre[w];
  }
}
int ans[11], save[11];
void merge_ans(int a, int b, int c) {
  int i, l = 1, r = 1, k = 1;
  memset(save, 0, sizeof(save));
  while (l <= ans[0] && r <= dp[a][b][0] && k <= c) {
    if (ans[l] <= dp[a][b][r]) {
      save[k++] = ans[l];
      l++;
    } else {
      save[k++] = dp[a][b][r];
      r++;
    }
  }
  while (l <= ans[0] && k <= c) {
    save[k++] = ans[l];
    l++;
  }
  while (r <= dp[a][b][0] && k <= c) {
    save[k++] = dp[a][b][r];
    r++;
  }
  save[0] = k - 1;
  for (i = 0; i < 11; i++) ans[i] = save[i];
}
void solve(int a, int b, int c) {
  int i, h;
  if (de[a] < de[b]) {
    h = a;
    a = b;
    b = h;
  }
  for (i = 19; i > 0; i--)
    if (de[a] - (1 << (i - 1)) >= de[b]) {
      merge_ans(a, i, c);
      a = fa[a][i];
    }
  if (a == b) {
    merge_ans(a, 1, c);
    return;
  }
  for (i = 19; i > 0; i--)
    if ((de[a] - (1 << (i - 1)) >= 0) && fa[a][i] != fa[b][i]) {
      merge_ans(a, i, c);
      merge_ans(b, i, c);
      a = fa[a][i];
      b = fa[b][i];
    }
  merge_ans(a, 1, c);
  merge_ans(b, 1, c);
  merge_ans(fa[a][1], 1, c);
}
int main() {
  int a, b, c, i, n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  tot = 0;
  memset(u, -1, sizeof(u));
  for (i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    add_b(a, b);
    add_b(b, a);
  }
  memset(dp, 0, sizeof(dp));
  for (i = 1; i <= m; i++) {
    scanf("%d", &a);
    add_p(a, i);
  }
  memset(fa, 0, sizeof(fa));
  memset(de, 0, sizeof(de));
  dfs(1, 1);
  while (q--) {
    scanf("%d%d%d", &a, &b, &c);
    memset(ans, 0, sizeof(ans));
    solve(a, b, c);
    printf("%d ", ans[0]);
    for (i = 1; i <= ans[0]; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}